I'm getting the hang of induction proofs but just can't seem to complete the inductive step. Meanwhile i beleive what i have made so far is correct.
I have a quantity : $n∈N$
And a statement:$$\sum\limits_{i=0}^n 3^i = \frac{1}{2}(3^{n + 1} - 1)$$
I perform the basis step by assuming that $P(0)$ which gives the following:
$$\sum\limits_{i=0}^0 3^i = \frac{1}{2}(3^{0 + 1} - 1)$$ This proves to be true. So assuming that this is true i can make the assumption of induction since it should be true for all arbitrary numbers, the assumption is as follows:
$$1 + 2 + ... + k = \sum\limits_{i=0}^k 3^i = \frac{1}{2}(3^{k + 1} - 1)$$
In the induction step we prove that if P(k) is true then $ P(k) \xrightarrow{} P(k + 1)$
With this assumption we would like to show that:
$$1 + 2 + ... + k + (k + 1) = \sum\limits_{i=0}^{k + 1} 3^i = \frac{1}{2}(3^{(k + 1) + 1} - 1)$$
We use the assumption of induction to create a statement which we would like to prove this is as follows:
$$\sum\limits_{i=0}^{k + 1} 3^i = \bigg ( \sum\limits_{i=0}^{k} 3^i\bigg) + 3^{i + 1} = \frac{1}{2}(3^{(k + 1)+1} - 1)$$
I start solving: $$1 + 2 + ... + k + (k + 1) = (1 + 2 + ... + k) + (k + 1)\\ = \frac{1}{2}(3^{k + 1} - 1) + (k + 1)\\ = \frac{3^{k + 1}-1}{2} +(k + 1)\\ = \frac{3^{k + 1} - 1 + 2(k + 1)}{2}\\ = \text{Result:} = \frac{1}{2}(3^{(k + 1) + 1} - 1)$$
I just can't get to the result which shows that my inductive assumption is true, and can't seem to figure out if i did something wrong or, just can't see forest from trees... Thanks for all the help. It has be solved by induction.
the formula is true for $n=0$.
Now suppose that $P(n)=\frac{1}{2}(3^{n+1}-1)$ for some $n \in \mathbb N \cup \{0\}$.
With the aid of 2. you have to show that $P(n+1)=\frac{1}{2}(3^{n+2}-1)$:
Since $P(n+1)=P(n)+3^{n+1}$, we get by 2. :
$P(n+1)=\frac{1}{2}(3^{n+1}-1)+3^{n+1}$.
It is your turn to show that
$$\frac{1}{2}(3^{n+1}-1)+3^{n+1}=\frac{1}{2}(3^{n+2}-1).$$